# What's the physical significance of using fourier transform for diffraction?

I am studying some basic idea of diffraction and there mention in far field, the diffraction pattern could be understood by Fourier transform. But I just don't understand what's the physical fact for that. And why cannot use Fourier transform for the the near-field case?

Also, when I am trying to understand the theory of diffraction, it ends up with some complicate math (integrals). I want to learn that but the books I am reading are not easy to understand. Anyone recommends some good books or video lectures (more theoretical but to explain most of the math in plain way)?

• I think that Fourier transform can be used for the far field (Fraunhofer diffraction) and the near case (Fresnel), but it's much more complicated. What math level do you have? May 12, 2013 at 8:36
• I wrote an answer about that here a while ago... May 12, 2013 at 15:11
• @jinawee You are right, that the concept of Fourier transformation requieres a university math level. However the idea that this transformation allows easier understanding of periodic events. Such as in acoustics, where a note C shows a peak at $440\,$Hz in fourier/frequency spectrum. May 17, 2013 at 17:43

To give an unmathematical catchy answer, let's look at Fraunhofer diffraction in double slit experiment. Interference at the observation plane depends on slit parameter $d$. What is the frequency of slits? E.g. $1\,\text{mm}\frac{1}{d}$: number of slits per length. Concluding frequiency in the setup. The following argumentation links this frequency to the fourier transform. The physical significance is in the real optics setup. The setup is easier described, when transformed in fourier space.
Using trigonometry first compute phase difference $\Delta\phi(\theta,d)$. Go deeper in this concept using a sketch to visualize phase difference of $n\cdot\lambda$, $n\in\mathbf{N}$ as bright maxima in diffraction pattern. There is no magic in the next step. It's just a another point of view: Try to grasp $\frac{1}{d}$ as a parameter on its own: $\Delta\phi(\theta,\frac{1}{d})$.